A certain circle can be represented by the following equation. $x^2+y^2-4x-12y-9=0$ What is the center of this circle ? $($
Answer: The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2-4x-12y-9&=0\\\\ x^2+y^2-4x-12y&=9\\\\ (x^2-4x)+(y^2-12y)&=9 \text{(rearrange terms)}\\\\ (x^2-4x{+4})+(y^2-12y{+36})&=9{+4}{+36}\end{aligned}$ Notice that we must add ${4}$ and ${36}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 4 and 36?] Writing the equation in standard form $\begin{aligned}(x^2-4x{+4})+(y^2-12y{+36})&=9{+4}{+36}\\\\ (x-2)^2+(y-6)^2&=49\\\\ (x-2)^2+(y-6)^2&=7^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(2,6)$ and has a radius of $7$ units. Summary The circle is centered at $(2,6)$. The circle has a radius of $7$ units.